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HPCL Electrical Engineer Mock Test - 2 - PSSSB Clerk MCQ


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30 Questions MCQ Test - HPCL Electrical Engineer Mock Test - 2

HPCL Electrical Engineer Mock Test - 2 for PSSSB Clerk 2024 is part of PSSSB Clerk preparation. The HPCL Electrical Engineer Mock Test - 2 questions and answers have been prepared according to the PSSSB Clerk exam syllabus.The HPCL Electrical Engineer Mock Test - 2 MCQs are made for PSSSB Clerk 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for HPCL Electrical Engineer Mock Test - 2 below.
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HPCL Electrical Engineer Mock Test - 2 - Question 1

A person has Rs. 5,350 in the form of denomination of Rs. 50, Rs. 100 and Rs. 500 and the total number of notes with him is 32. Find how many notes of Rs. 50, Rs.100 and Rs. 500 denominations, respectively, he has with him?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 1

Calculation:

Let x, y, and z be the number of Rs. 50, Rs. 100, and Rs. 500 notes, respectively.

According to the question,

x + y + z = 32 and 50x + 100y + 500z = 5350.

First, let's eliminate z from the second equation by substituting z = 32 - x - y

⇒ 50x + 100y + 500(32 - x - y) = 5350

Simplify the equation:

50x + 100y + 16000 - 500x - 500y = 5350

⇒ -450x - 400y = -10650 (Divide by -50)

9x + 8y = 213

Now, we can find a solution by trial method:

x = 13, y = 12 → 9(13) + 8(12) = 213 (This solution works)

So, x = 13 and y = 12. Now, we can find z:

z = 32 - x - y = 32 - 13 - 12 = 7

∴ The correct answer is x=13 (Rs. 50 notes), y=12 (Rs. 100 notes), and z=7 (Rs. 500 notes).

HPCL Electrical Engineer Mock Test - 2 - Question 2

How many spherical lead shots each of radius 2.1 cm can be obtained from a solid rectangular lead piece with dimensions 84 cm, 77 cm and 63 cm.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 2

Given :

Spherical lead shots of radius 2.1 cm.

Rectangular lead piece of dimension 84 cm, 77 cm, 63 cm.

Formula used :

Volume of sphere = 4/3πr3

Volume of cuboid = l × b × h

Calculation :

According to question,

Let the number of spherical lead shots be n.

⇒ 84 × 77 × 63 = n × 4/3 × 22/7 × 2.1 × 2.1 × 2.1

⇒ 21 × 77 × 63= n × 1/3 × 22/7 × 21/10 × 21/10 × 21/10

⇒ 21 × 77 × 63 = n × 22 × 21/10 × 21/10 × 1/10

⇒ 77 × 63 = n × 22 × 21/1000

⇒ 7/2 × 63 = n × 21/1000

⇒ n = 7/2 × 3000

⇒ n = 7 × 1500

⇒ n = 10500

∴ The correct answer is 10500.

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HPCL Electrical Engineer Mock Test - 2 - Question 3

A candidate who gets 50 percent marks fails by 100 marks but another candidate who got 78 percent marks gets 30 percent more than the passing marks. The maximum marks is:

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 3

Given:

Candidate A's score = 50% and is short by 100 marks.

Candidate B's score = 78% and gets 30% more than passing marks.

Concept:

Use the concept of percentages and linear equations.

Solution:

A's score = 50% of total marks + 100 = Passing marks

B's score = 78% of total marks - 30% of passing marks

Equating both we get:

50% of total marks + 100 = 78% of total marks - 30% of passing marks

⇒ 50% of total marks + 100 = 78% of total marks - 30%(50% of total marks + 100)

⇒ 1/2 of total marks + 100 = 78/100 of total marks - 30/100(1/2 of total marks + 100)

⇒ 1/2 of total marks + 100 = 78/100 of total marks - 15/100 of total marks - 30

⇒ 13/100 of total marks = 130

⇒ Total marks = 1000

Therefore, the maximum marks are 1000.

HPCL Electrical Engineer Mock Test - 2 - Question 4
The areas of three adjacent faces of a cuboid are 192 cm2, 144 cm2 and 108 cm2. Two-third of the volume of this cuboid is equal to :
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 4

Given:

The areas of three adjacent faces are 108 cm2, 144 cm2 and 192 cm2

Formula Used:

Volume of cuboid = lbh

l = length, b = breadth and h = height

Calculation:

According to the question

lb = 108 -----(i)

bh = 144 -----(ii)

hl = 192 -----(iii)

On multiplying (i), (ii) and (iii), we get

lb × bh × hl = 108 × 144 × 192

⇒ (lbh)2 = 108 × 144 × 192

⇒ (Volume of cuboid)2 = 36 × 3 × 144 × 3 × 64

⇒ (Volume of cuboid)2 = (6 × 3 × 12 × 8)2

⇒ Volume of cuboid = 1728 cm3

According to the question

Volume of cuboid = 2/3 × 1728 = 1152 cm3

∴ The required answer will be 1152 cm3

HPCL Electrical Engineer Mock Test - 2 - Question 5

Choose the correct alternative from given ones that will complete the series :

1250, 250, 50, 10 , ?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 5

To complete the series: 1250, 250, 50, 10, ?

The pattern involves dividing each number by 5 to obtain the next:

1250 / 5 = 250

250 / 5 = 50

50 / 5 = 10

10 / 5 = 2

So, the number that completes the series is 2.

HPCL Electrical Engineer Mock Test - 2 - Question 6

Find the missing number in the series given below.

2, 3, 5, 7, 11, 13, (?)

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 6

Logic: all the given numbers are prime number.

So, next number will be a prime number and after 13 next prime number is 17.

Hence, the next term is 17.

HPCL Electrical Engineer Mock Test - 2 - Question 7

Fill in the blank with the appropriate form of the Verb that agrees with the Subject.

Each one of the workers _______ a raise in salary from the coming month.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 7

The correct answer is Option (2) i.e. "wants".

Key Points

  • Here "Each one" is the main subject which is singular.
  • "Each one of the workers" implies each worker individually which requires a singular verb.
  • "Wants" correctly agrees with the singular subject phrase.

Therefore, the correct answer is- "Each one of the workers wants a raise in salary from the coming month."

Additional Information

  • Option (1) "are wanting" does not agree with the singular subject phrase.
  • Option (3) "have wanted" does not correctly fit the singular subject in present tense.
  • Option (4) "want" is plural and does not agree with the singular subject phrase.
HPCL Electrical Engineer Mock Test - 2 - Question 8

Select the word segment that substitutes (replaces) the bracketed word segment correctly and completes the sentence meaningfully. If there is no need to substitute it, select "No Correction"

Stonehenge (have capture) the hearts of people from pre-historic times due to its unique structure.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 8
The correct answer is 'has captured'.
Key Points
  • The subject of the sentence, 'Stonehenge', is singular, and it requires the use of 'has' instead of 'have'.
  • In addition to the appropriate auxiliary verb 'has', the main verb 'capture' needs to be in the past participle form 'captured' to fit properly within the present perfect tense structure 'has captured'.
  • The present perfect tense 'has captured' correctly implies an action that happened at an unspecified time before now, which fits the context of the sentence.

​Therefore, the correct answer is 'Option 3'.

Corrected Sentence: Stonehenge has captured the hearts of people from pre-historic times due to its unique structure.

Additional Information
  • 'Have capture' (option 1) and 'No correction' (option 4) - Both of these options keep the grammatically incorrect form found in the original statement, either by maintaining the incorrect auxiliary verb 'have' with a singular subject or by not changing the base form of the verb 'capture'.
  • 'Have captured' (option 2) - Although 'captured' is the correct form of the main verb to use here, the auxiliary verb 'have' does not agree in number with the singular subject 'Stonehenge', making this option incorrect as well.
HPCL Electrical Engineer Mock Test - 2 - Question 9

The perimeter of the top of a rectangular table is 30 m, whereas its area is 54 m2. What is the length of its diagonal?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 9

As we see in the figure,

Let, length of the rectangular table = x m.

Breadth of the rectangular table = y m.

According to problem,

⇒ 2(x + y) = 30 ………..

⇒ x + y = 15 ….. (1)

According to problem,

⇒ xy = 54 …. (2)

From (1) and (2) we get,

⇒ x = 9 and y = 6

∴ Length of the diagonal,

⇒ √(92 + 62)

⇒ √(81 + 36)

⇒ √117

⇒ 10.82 m.

HPCL Electrical Engineer Mock Test - 2 - Question 10
Rahul sells two helmets at the rate of Rs. 1232 each. He gains 12% on one and loses 12% on the other. What will be the total loss (in Rs) in the whole transaction?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 10

Calculation:

SP of one helmet = 1232

SP of two helmets = 1232 × 2 = 2464

CP of first helmet = 1232 × 100/112 = 1100

CP of second helmet = 1232 × 100/88 = 1400

Total CP of two helmets = 1100 + 1400 = 2500

As we know, loss = CP - SP

∴ Loss = 2500 - 2464 = 36
HPCL Electrical Engineer Mock Test - 2 - Question 11
M is 25 per cent more than N. N is how much percent less than M?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 11

Given:

M is 25% more than N.

Concept Used:

If A is R% more than b,

Then B is [R/(100 + R) × 100]% less than A less than A.

Calculation:

N = [R/(100 + R) × 100]%

⇒ [25/(100 + 25) × 100]%

⇒ [25/125 × 100]%

⇒ 1/5 × 100]% = 20%

HPCL Electrical Engineer Mock Test - 2 - Question 12
The compound interest (compounding half yearly) received on Rs. 10000 for 2 years is Rs. 10736. What is the rate of interest per annum?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 12

Given:

Compound interest = Rs. 10736

Principal = Rs. 10000

Time = 2 years

Formula used:

Amount when interest is compounded half-yearly = P[1 + (R/200)]2T

P = Principal

R = Rate of interest

T = Time taken

Calculation:

According to the question,

Amount = 10000 + 10736 = 20736

20736 = 10000[1 + (R/200)]4

⇒ 20736/10000 = [1 + (R/200)]4

⇒ (12/10) - 1 = R/200

⇒ R = 40

∴ Rate of interest = 40%

HPCL Electrical Engineer Mock Test - 2 - Question 13

A sum of money becomes Rs. 39930 in 3 years and Rs. 36300 in 2 years when kept at compound interest (compounding annually). What is the sum?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 13

Given:

Amount in 3 years = Rs. 39930

Amount in 2 years = Rs. 36300

Formula used:

A = P(1 + r/100)n

Where, A = Amount, r = Rate of interest, n = Time in year

Calculation:

According to the question,

39930 = P(1 + r/100)3 ----(1)

36300 = P(1 + r/100)2 ----(2)

Dividing equation (1) by equation (2), we get

(39930/36300) = [(1 + r/100)3/(1 + r/100)2]

⇒ 11/10 = 1 + r/100

⇒ r/100 = 1/10

⇒ r = 10%

From equation (2),

36300 = P[1 + (10/100)]2

⇒ 36300 = P(11/10)2

⇒ P = 36300 × (100/121)

⇒ P = 30000

∴ The sum is Rs. 30000.

HPCL Electrical Engineer Mock Test - 2 - Question 14
14 men complete a work in 18 days. If 21 men are employed, then the time required to complete the same work will be:
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 14

Given:

14 men complete a work in 18 days.

Calculation:

According to the question

14 men complete a work = (14 × 18) days

Then,

21 men complete the same work = [(14 × 18)/21] days

⇒ (2 × 6) days

⇒ 12 days

∴ The time required to complete the same work will be 12 days

HPCL Electrical Engineer Mock Test - 2 - Question 15
If Two sixth of a number is 80 more than 25% of the same number, Find the number.
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 15

Given:

Two-sixths of a number is 80 more than 25% of the same number

Calculation:

Let the number be x.

⇒ x × (2/6) = x × (25/100) + 80

⇒ x × (1/3) - x × (1/4) = 80

⇒ x × (1/12) = 80

⇒ x = 960

The number is 960.

HPCL Electrical Engineer Mock Test - 2 - Question 16
What is the greatest 5-digit number which on being divided by 3, 5, 6 and 9 leaves remainders 1, 3, 4 and 7, respectively?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 16

Given:

The greatest 5-digit number which on being divided by 3, 5, 6 and 9 leaves remainders 1, 3, 4 and 7, respectively

Solution:

LCM of 3, 5, 6 and 9 = 90

We Know that the greatest 5 digit number = 99999

The greatest 5 digit number which is divisible by 90 = 99990

To get remainders 1, 3, 4, 7 when divided by 3, 5, 6, 9 respectively we find difference of these umbers, As

⇒ 3 - 1 = 2

5 - 3 = 2

⇒ 6 - 4 = 2

9 - 7 = 2

Therefore, The greatest 5-digit number which on being divided by 3, 5, 6 and 9 leaves remainders 1, 3, 4 and 7, respectively

⇒ 99990 - 2 = 99988

HPCL Electrical Engineer Mock Test - 2 - Question 17
If the 5-digit number 9x34y is divisible by 24, then what is the maximum value of (x + y)?
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 17

Given:

Five digit number is given = 9x34y, which is divisible by 24.

Concept used:

Divisibility rule of 8 and 3 which is factor of 24.

Divisibility rule of 8 is last three digit of number is divisible by 8.

Divisibility rule of 3 is sum of all digits of given number is divisible by 3.

Calculation:

The number is = 9x34y

First we check divisibility rule of 8,

According to this rule, last three digits is divisible by 8,

⇒ 34y is divisible by 8,

34y / 8

When we divide 34y by 8 ,for complete divisibisible by 8 y should be 4 344 / 8 (Let y = 4)

Check the divisibility rule of 3,

According to this rule, sum of all digits is divisible by 3.

9 + x + 3 + 4 + y = 16 + x + y is divisible by 3.

Put y = 4,

16 + x + 4 = 20 + x

Which is completely divisible by 3 when x = 1 and 7

We take maximum value because question ask maximum value, x = 7

20 + 7 = 27 is divisible by 3.

X = 7 and y = 4

x + y = 7 + 4 = 11

∴Option 1 is correct.

HPCL Electrical Engineer Mock Test - 2 - Question 18

In a certain code language, ‘EDUCATION’ is written as ‘GHFXDLWQR’. How will ‘FRAGRANCE’ be coded as in that language?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 18

The logic followed here is :-

Similarly,

Hence, the code for "FRAGRANCE' will be 'UIJDUQDHF'.

HPCL Electrical Engineer Mock Test - 2 - Question 19

Read the given statements and conclusions carefully. Decide which of the given conclusions logically follows from the statements.

Statement: “It is difficult to form true selfless friendship with women after their marriage because they always give priority to their family over their friends.” – A woman complains to her husband.

Conclusion:

I. Families should be given priority by the elders to maintain good relations within the family.

II. Friendship with a married person is not possible.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 19

Given:

Statements: “It is difficult to form true selfless friendship with women after their marriage because they always give priority to their family over their friends.” – A woman complains to her husband.

Conclusion:

I. Families should be given priority by the elders to maintain good relations within the family → False (Because it is given in the statement that women always give priority after their marriage to their family over their friends and there is no such information mentioned regarding elders.)

II. Friendship with a married person is not possible → False (Because no such information is given regarding married person.)

Hence, the correct answer is "Option 4".

HPCL Electrical Engineer Mock Test - 2 - Question 20

In the following figure, triangle represents personal loan, circle represents private bank, hexagon represents government bank, rectangle represents education loan and square represents housing loan.

What does letter ‘L’ represents?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 20

As per the given information,

The shaded-by part i.e. L represents → Private banks which provide education loan and personal loan but not housing loan

Hence, the correct answer is "option (1)".

HPCL Electrical Engineer Mock Test - 2 - Question 21

Two wattmeters are used to measure the power in a 3-phase balanced system. What is the power factor of the load when one wattmeter reads twice the other?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 21

Concept:

In a two-wattmeter method, for lagging load

The reading of first wattmeter (W1) = VL IL cos (30 + ϕ)

The reading of second wattmeter (W2) = VL IL cos (30 - ϕ)

Total power in the circuit (P) = W1 + W2

Total reactive power in the circuit

Power factor = cos ϕ

Calculation:

Given that, W1 = 2 W2

⇒ ϕ = 30°

Power factor = cos 30° = 0.866

Important Point:

HPCL Electrical Engineer Mock Test - 2 - Question 22

For the circuit shown, the value of R is so adjusted as to transfer the maximum power in 1 Ω resistor. What is the amount of power?

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 22

At R = 0 Ω, the total supply voltage will be applied across the load resistor of 1 Ω.

Thus, at R = 0 Ω, maximum power will occur in the 1 Ω resistor.

At this case:

HPCL Electrical Engineer Mock Test - 2 - Question 23
A circular loop has its radius increasing at a rate of 1 m/s. The loop placed perpendicular to a constant magnetic field of 0.8 wb/m2. When the radius of the loop is 2 m, the emf induced in the will be
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 23

Given that,

Rate of radius (dr/dt) = 1 m/s

Magnetic field (B) = 0.8 wb/m2

Radius of the loop (r) = 2 m

Emf induced

= 2π × 2 × 0.8 × 1 = 3.2π V

HPCL Electrical Engineer Mock Test - 2 - Question 24

The Buchholz relay is normally used to protect the

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 24

  • Buchholz relay is a protection device which is normally used in large oil absorbed transformers; It is a kind of oil and gas activated relay
  • Whenever a small fault happens within the electrical device, heat is generated by the fault currents
  • The generated heat causes decomposition of electrical device oil and gas bubbles are generated
  • These gas bubbles run in the upward direction and obtain collected within the Buchholz relay
  • The collected gas relocates the oil in Buchholz relay and therefore the displacement is similar to the amount of gas collected
  • The dislocation of oil causes the higher float to shut the higher mercury switch to connect an alarm circuit
  • Hence, once a small fault happens, then the alarm will be activated; The collected quantity of gas specifies the level of fault occurred
  • Different types of faults have different types of oil flow and thus a fault can be recognized easily, and proper action can be taken
  • It is used for the transformers of rating greater than 500 KVA.
HPCL Electrical Engineer Mock Test - 2 - Question 25
A single-phase energy meter is operating on 200 V, 50 Hz supply with a load of 10 A for two hours at 0.8 p.f. The meter takes 1800 revolutions in that period. The meter constant is:
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 25

Energy meter:

An energy meter or watt-hour meter is used to measure the energy consumed in kWh.

It is an integrating type instrument.

Recording mechanism:

It is required for the energy meter to record the no. of revolutions made by the aluminum disc which is proportional to the energy consumed in kWh.

The meter constant can be calculated as

k = (No. of revolutions / Energy consumption in kWh)

Units of meter constant are revolutions per kWh

1 unit means 1 kWh.

Calculation:

Given:

Voltage V= 200 V

Current I = 10 A

Time = 2 hour

cosϕ = 0.8

Energy consumed = V × I × cosϕ × time = 200 × 10 × 0.8 × 2 = 3.2 kWh

Meter constant = No. of revolution by meter/Energy consumed = 1800/3.2 = 562.5 rev/kWh.

Points to remember:

Braking torque obtained by a permanent magnet inside the meter used to control the speed of the disc.

HPCL Electrical Engineer Mock Test - 2 - Question 26

A closed surface encloses the point charge Q and has a medium with permittivity of free space ε0. Then the Gauss’s law should not be represented by

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 26

Gauss law for electric field:

Gauss's law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

The net flux of a given electric field through a given surface, divided by the enclosed charge should be equal to a constant.

Integral form:

Differential form: ∇ . E = ρ/ε0

Gauss law for magnetic field:

Gauss's law for magnetism is one of the four Maxwell's equations. It states that the magnetic field B has divergence equal to zero, in other words, that it is a solenoidal vector field.

It is equivalent to the statement that magnetic monopoles do not exist.

Integral form:

Differential form: ∇ . B = 0

Gauss law can’t be represented by:

HPCL Electrical Engineer Mock Test - 2 - Question 27
For maximum current during 'Slip Test' on a synchronous machine, the armature mmf aligns along -
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 27

Concept:

Slip test:

Direct and quadrature axis reactance can be determined by this test.

During this test, the synchronous machine is driven by the separate prime mover at slightly different from synchronous speed.

The field winding is open-circuited and rated frequency and reduced voltage is applied across the terminal.

Under these conditions, the relative velocity between the field pole and the rotating armature m.m.f wave is equal to the difference between synchronous speed and rotor speed.

Explanation:

For maximum current, during slip test on a Synchronous machine, the armature MMF aligns along the q-axis.

HPCL Electrical Engineer Mock Test - 2 - Question 28

In the medium transmission line, the shunt admittance is placed in the middle and the series impedance is divided into two equal parts which are placed on either side of the shunt admittance. From the given context, analyze the representation of the medium transmission line.

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 28

Nominal T Model Of A Transmission Line:

In a nominal T model of a medium transmission line, the series impedance is divided into two equal parts, while the shunt admittance is concentrated at the middle of the line. The nominal T model of a medium transmission line is shown in the figure.

Sending end voltage and current can be obtained by application of KVL and KCL, to the circuit shown below

Current in the capacitor can be given as,

By Kirchhoff’s current law at node a,

By Kirchhoff’s voltage law

Equation of Sending end voltage Vs and current Is can be written in the matrix form as

Also,

Hence, the ABCD constant of the nominal T-circuit model of a medium line is

HPCL Electrical Engineer Mock Test - 2 - Question 29

The magnitude of induced e.m.f is directly proportional to the

Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 29

Faraday's laws: Faradays performed many experiments and gives some law about electromagnetism.

  • Faraday's First Law: Whenever a conductor is placed in a varying magnetic field an EMF gets induced across the conductor (called as induced emf), and if the conductor is a closed circuit then induced current flows through it.
    A magnetic field can be varied by various methods –
    • By moving magnet
    • By moving the coil
    • By rotating the coil relative to a magnetic field
  • Faraday's second law of electromagnetic induction states that the magnitude of induced emf is equal to the rate of change of flux linkages with the coil.
  • According to Faraday's law of electromagnetic induction, the rate of change of flux linkages is equal to the induced emf

  • A magnet is moved towards the coil - Correct
  • A magnet is moved away from the coil - Correct
  • A magnet is kept stationary near the coil - Incorrect
HPCL Electrical Engineer Mock Test - 2 - Question 30
Total harmonic distortion is defined as:
Detailed Solution for HPCL Electrical Engineer Mock Test - 2 - Question 30

Total harmonic distortion or THD is a common measurement of the level of harmonic distortion present in power systems.

THD can be related to either current harmonics or voltage harmonics.

It is defined as the ratio of rms value of all the harmonic components to the rms value of the fundamental component.

Mathematically, it can be represented as

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